2. JK9972397103
An algorithm is a series of well defined steps which gives a
procedure for solving a type of problem.
A lemma is a proven statement used for proving another
statement.
The word algorithm comes from the name of the 9th century
Persian mathematician al-Khwarizmi. In fact, even the word
‘algebra’is derived from a book, he wrote, called Hisab
al-jabr w’al-muqabala.
Euclid’s division algorithm,as the name suggest,has to do with divisibility of integers.It
says any positive integer “a” can be divided by any another positive integer “b” in such a
way that it leaves a remainder “r” that is smaller than “b”.
EUCLID’S DIVISION LEMMA(THEOREM NO:-1.1)
Given positive integers “a” and “b” ,there exist unique integers “q” and “r” satiafying
a=bq+r,0≤r<b.
3. JK9972397103
EXERCISE NO: 1.1
1. Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
135 and 225
Since 225 > 135, we apply the division lemma to 225 and 135 to
obtain
225 = 135 × 1 + 90
Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to
obtain
135 = 90 × 1 + 45
We consider the new divisor 90 and new remainder 45, and apply the
division lemma to obtain
90 = 2 × 45 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45,
Therefore, the HCF of 135 and 225 is 45.
4. JK9972397103
(ii)196 and 38220
Since 38220 > 196, we apply the division lemma to 38220 and 196 to
obtain
38220 = 196 × 195 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.
(iii)867 and 255
Since 867 > 255, we apply the division lemma to 867 and 255 to
obtain 867 = 255 × 3 + 102
Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102
to obtain 255 = 102 × 2 + 51
We consider the new divisor 102 and new remainder 51, and apply the
division lemma to obtain 102 = 51 × 2 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 51,
Therefore, HCF of 867 and 255 is 51.
5. JK9972397103
2.Show that any positive odd integer is of the form, 6q + 1 or 6q + 3, or 6q + 5 , where q is
some integer.
Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,
a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r< 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an Integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an Integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.
Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5
6. JK9972397103
3.An army contingent of 616 members is to march behind an army band of 32 members
in a parade. The two groups are to march in the same number of columns. What is the
maximum number of columns in which they can march?
HCF (616, 32) will give the maximum number of columns in which they can march.
We can use Euclid’s algorithm to find the HCF.
616 = 32 × 19 + 8
32 = 8 × 4 + 0
The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.
7. JK9972397103
4.Use Euclid’s division lemma to show that the square of any positive integer is either of form
3m or 3m + 1 for some integer m. [Hint: Let x be any positive integer then it is of the form 3q,
3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form
3m or 3m + 1.]
Let a be any positive integer and b = 3.
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
Or,
a2 = (3q)2 or (3q+1)2 or (3q+2)2
a2 = (9q2) or (9q2 + 6q + 1) or (9q2 +12q+4)
= 3 x (3q2) or 3 x (3q2+2q) + 1 or 3 x (3q2+4q+1) + 1
= 3k1 or 3k2+1 or 3k3 +1
Where k1, k2, and k3 are some positive integers
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.
8. JK9972397103
5.Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m,
9m + 1 or 9m + 8.
Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
∴a=3q or 3q+1 or 3q+2
Therefore, every number can be represented as these three forms.
There are three cases.
Case 1: When a = 3q,
a 3=(3q)3= 27q3 =9(3q3)= 9m, Where m is an integer such that m=3q3
Case 2: When a = 3q + 1, a3 = (3q +1)3
a3 = 27q3 + 27q2 + 9q + 1 a3= 9(3q3 + 3q2 + q) + 1
a3 = 9m + 1, Where m is an integer such that m = (3q3 + 3q2 + q)
Case 3: When a = 3q + 2, a3 = (3q +2)3
a3 = 27q3 + 54q2 + 36q + 8 a3= 9(3q3 + 6q2 + 4q) + 8
a3 = 9m + 8, Where m is an integer such that m = (3q3 + 6q2 + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.
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THE FUNDAMENTAL THEOREM OF ARITHMETIC
(THEOREM NO:-1.2)
An equivalent version of Theorem 8.2 was probably first
recorded as Proposition 14 of Book IX in Euclid’s Elements,
before it came to be known as the Fundamental Theorem
of Arithmetic. However, the first correct proof was given by
Carl Friedrich Gauss in his Disquisitiones Arithmeticae.
Every composite number can be expressed (factorized)as a product of
primes,and this factorization is unique,apart from the order in which the
prime factors occur.
The Fundamental Theorem of Arithmetic says that every composite number can be factorised as a
product of primes. Actually it says more. It says that given any composite number it can be
factorised as a product of prime numbers in a ‘unique’ way, except for the order in which the
primes occur. That is, given any composite number there is one and only one way to write it as a
product of primes,as long as we are not particular about the order in which the primes occur. So,
for example, we regard 2 × 3 × 5 × 7 as the same as 3 × 5 × 7 × 2, or any other possible order in
which these primes are written.
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4.Given that HCF (306, 657) = 9, find LCM (306, 657).
HCF (306, 657) = 9
We know that, LCM × HCF = Product of two numbers
∴LCM × HCF = 306 × 657
LCM =
306 × 657
HCF
=
306 × 657
9
LCM = 22338
5.Check whether 6n can end with the digit 0 for any natural number n.
If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be
divisible by 2 and 5 as 10 = 2 × 5
Prime factorisation of 6n = (2 ×3)n
It can be observed that 5 is not in the prime factorisation of 6n.
Hence, for any value of n, 6n will not be divisible by 5.
Therefore, 6n cannot end with the digit 0 for any natural number n.
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6.Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Numbers are of two types - prime and composite. A prime number can be divided by 1 and
only itself,
where as a composite number have factors other than 1 and itself. It can be observed that
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1)
= 13 × 78
= 13 ×13 × 6
The given expression has 6 and 13 as its factors. Therefore, it is a composite number.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 ×1009
1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its
factors.
Hence, it is a composite number.
16. JK9972397103
7.There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of
the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point
and at the same time, and go in the same direction. After how many minutes will they meet
again at the starting point?
It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular
path. As they are going in the same direction, they will meet again at the same time when Ravi
will have completed 1 round of that circular path with respect to Sonia. And the total time
taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and
Ravi for completing 1 round of circular path respectively
i.e., LCM of 18 minutes and 12 minutes.
18 = 2 ×3 ×3
And, 12 = 2 ×2 ×3
LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36
Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.
17. JK9972397103
REVISITING IRRATIONAL NUMBERS
A number ‘s’ is called irrational if it cannot be written in the form
𝑝
𝑞
,where p and q are integers
and q ≠ 0.
The sum or difference of a rational and an irrational number is irrational.
The product and quotient of a non-zero rational and irrational number is irrational.
THEOREM NO:-1.3
Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer.
18. JK9972397103
EXERCISE NO: 1.3
1.Prove that 5 is irrational.
Let 5 is a rational number. Therefore, we can find two integers a, b (b ≠ 0) such that 5 =
𝑎
𝑏
Let a and b have a common factor other than 1.
Then we can divide them by the common factor, and assume that a and b are co-prime.
a = 5 b squaring on both sides,then a2=5b2⇒
a2
5
=b2
Therefore, a2 is divisible by 5 and it can be said that a is divisible by 5.⟶(1)
Let
a
5
=k,then a= 5k, where k is an integer.then we know that a2=5b2 by a value ,we will get
(5k)2=5b2 ⇒25 k2=5b2 ⇒k2=
b2
5
This means that b2 is divisible by 5 and hence, b is divisibleby 5. ⟶(2)
From (1) & (2), implies that a and b have 5 as a common factor. And this is a contradiction to
the fact that a and b are co-prime.
Hence, 5 cannot be expressed as
𝑝
𝑞
or it can be said that 5 is irrational.
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2.Prove that is irrational 3+2 5.
Let is 3+2 5 rational.
Therefore, we can find two integers a, b (b ≠ 0) such that 3+2 5 =
𝑎
𝑏
2 5 =
𝑎
𝑏
-3⇒ 5 =
1
2
𝑎
𝑏
−3
Since a and b are integers,
1
2
𝑎
𝑏
−3 will also be rational and therefore, 5 is rational.
This contradicts the fact that 5 is irrational. Hence, our assumption
that 3+2 5 is rational is false. Therefore, 3+2 5 is irrational.
3.Prove that the following are irrationals: (i)
1
2
Let
1
2
is rational.
Therefore, we can find two integers a, b (b ≠ 0) such that
1
2
=
𝑎
𝑏
⇒ 2=
𝑏
𝑎
But
𝑏
𝑎
is rational as a
and b are integers. Therefore, 2 is rational which contradicts to the fact that 2 is irrational.
Hence, our assumption is false and
1
2
is irrational.
20. JK9972397103
(ii) 7 5
Let 7 5 is rational.
Therefore, we can find two integers a, b (b ≠ 0) such that 7 5 =
𝑎
𝑏
5 =
𝑎
7𝑏
Since a and b are integers,
𝑎
7𝑏
will also be rational and therefore, 5 is rational.
This contradicts the fact that 5 is irrational. Hence, our assumption
that 7 5 is rational is false. Therefore, 7 5 is irrational.
(iii) 6+ 2.
Let is 6+ 2 rational.
Therefore, we can find two integers a, b (b ≠ 0) such that 6+ 2 =
𝑎
𝑏
2 =
𝑎
𝑏
-6,Since a and b are integers,
𝑎
𝑏
-6will also be rational and therefore, 2 is rational.
This contradicts the fact that 2 is irrational. Hence, our assumption
that 6+ 2 is rational is false. Therefore, 6+ 2 is irrational.
21. JK9972397103
REVISITING RATIONAL NUMBERS AND THEIR DECIMAL EXPANSIONS
THEOREM NO:-1.5
Let x be a rational number whose decimal expansion terminates.Then x can be expressed in the
form
𝑝
𝑞
,where p and q are coprime, and the prime factorisation of q is of the form 2n5m, where n,
m are non-negative integers.
THEOREM NO:-1.6
Let x =
𝑝
𝑞
be a rational number, such that the prime factorization of q is of the form 2n5m, where n,
m are non-negative integers. Then x has a decimal expansion which terminates.
THEOREM NO:-1.7
Let x =
𝑝
𝑞
be a rational number, such that the prime factorization of q is not of the form 2n5m ,where
n, m are non-negative integers. Then, x has a decimal expansion which is non-terminating
repeating (recurring).
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EXERCISE NO: 1.4
1. Without actually performing the long division, state whether the following rational numbers
will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
(i)
13
3125
3125 = 55
The denominator is of the form 5m.
Hence, the decimal expansion of
13
3125
is terminating.
(ii)
17
8
8=23
The denominator is of the form 2m.
Hence, the decimal expansion of
17
8
is terminating.
(iii)
64
455
455 = 5 × 7 × 13 ,Since the denominator is not in the form 2m× 5n, and it also contains
7 and 13 as its factors, its decimal expansion will be non-terminating repeating.
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(iv)
15
1600
1600 = 26 × 52
The denominator is of the form 2m× 5n.
Hence, the decimal expansion of
15
1600
is terminating.
(v)
29
343
343=73
The denominator is not in the form 2m× 5n, and it also contains 7 as its factor, its decimal
expansion will be non-terminating repeating.
(vi)
23
2352
The denominator is of the form 2m× 5n.
Hence, the decimal expansion of
23
2352 is terminating.
(vii)
129
225775
The denominator is not in the form 2m× 5n, and it also contains 7 as its factor, its decimal
expansion will be non-terminating repeating.
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(viii)
6
15
=
23
35
=
2
5
The denominator is of the form 5n.
Hence, the decimal expansion of
6
15
is terminating.
(ix)
35
50
=
75
105
=
7
25
The denominator is of the form 2m× 5n.Hence the decimal expansion of
35
50
is terminating.
(x)
77
210
=
711
307
=
11
30
30=235
The denominator is not in the form 2m× 5n, and it also contains 3 as its factor, its decimal
expansion will be non-terminating repeating.
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2.Write down the decimal expansions of those rational numbers in Question 1 above which
have terminating decimal expansions. (i)
13
3125
13
3125
=
13
2055 Multiplying and dividing by25
13
2055
25
25=
1325
205525=
1332
15525=
416
(10)5=0.00416
(ii)
17
8
17
8
=
17
2350 Multiplying and dividing by53
17
2350
53
53=
1753
235053=
17125
12353=
2125
(10)3=2.125
27. JK9972397103
3.The following real numbers have decimal expansions as given below. In each case, decide
whether they are rational or not. If they are rational, and of the form
𝑝
𝑞
, what can you say
about the prime factor of q?
(i) 43.123456789
Since this number has a terminating decimal expansion,
it is a rational number of the form
𝑝
𝑞
and q is of the form2m5n
i.e., the prime factors of q will be either 2 or 5 or both.
(ii) 0.120120012000120000 …
The decimal expansion is neither terminating nor recurring. Therefore, the given number is an
irrational number.
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(iii) 43.123456789
Since the decimal expansion is non-terminating recurring, the given number is a rational
number of the form and q is not of the form
𝑝
𝑞
and q is of the form2m5n i.e., the prime factors
of q will also have a factor other than 2 or 5.
